Local Leopoldt’s Problem
for Rings of Integers in Abelian
p-Extensions of Complete Discrete Valuation Fields
M. V. Bondarko
Received: April 12, 2000 Revised: December 8, 2000
Communicated by A. S. Merkurev
Abstract. Using the standard duality we construct a linear embedding of an associated module for a pair of ideals in an extension of a Dedekind ring into a tensor square of its fraction field. Using this map we investi- gate properties of the coefficient-wise multiplication on associated orders and modules of ideals. This technique allows to study the question of de- termining when the ring of integers is free over its associated order. We answer this question for an Abelian totally wildly ramified p-extension of complete discrete valuation fields whose different is generated by an element of the base field. We also determine when the ring of integers is free over a Hopf order as a Galois module.
1991 Mathematics Subject Classification: 11S15, 11S20, 11S31
Keywords and Phrases: Complete discrete valuation (local) fields, addi- tive Galois modules, formal groups, associated Galois modules
Introduction
0.1. Additive Galois modules and especially the ring of integers of local fields are considered from different viewpoints. Starting from H. Leopoldt [L] the ring of integers is studied as a module over its associated order. To be precise, if K is an extension of a local field k with Galois group being equal to G and OK is the ring of integers ofK, thenOK is considered as a module over AK/k(OK) ={λ∈k[G], λOK ⊂OK}.
One of the main questions is to determine whenOK is free as anAK/k(OK)- module. Another related problem is to describe explicitly the ringAK/k(OK) (cf. [Fr], [Chi], [CM]).
This question was actively studied by F. Bertrandias and M.-J. Ferton (cf. [Be], [B-F], [F1-2]) and more recently by M. J. Taylor, N. Byott and G. Lettl (cf.
[T1], [By], [Le1]).
In particular, G. Lettl proved that ifK/Qp is Abelian andk⊂K, thenOK ≈ AK/k(OK) (cf. [Le1]). The proof was based on the fact that all Abelian extensions ofQp are cyclotomic. So the methods of that paper and of most of preceding ones are scarcely applicable in more general situations.
M. J. Taylor [T1] considers intermediate extensions in the tower of Lubin-Tate extensions. He proves that for some of these extensionsOKis a freeAK/k(OK)- module. Taylor considers a formal Lubin-Tate groupF(X, Y) over the ring of integerso of a local fieldk. Let π be a prime element of the field k and Tm
be equal to Ker[πm] in the algebraic closure of the fieldk. For 1≤r≤m, let L be equal to k(Tm+r), and let K be equal to k(Tm), . Lastly, let q be the cardinality of the residue field ofk.
Taylor proves that
(1) the ringOL is a freeAL/K(OL)-module and any element ofLwhose valu- ation is equal toqr−1 generates it, and
(2) AL/K(OL) = OK +Pqr−2
i=0 OKσi, where σi ∈ K[G] and are described explicitly (cf. the details in [T1], subsection 1.4).
This result was generalized to relative formal Lubin-Tate groups in the papers [Ch] and [Im]. Results of these papers were proved by direct computation. So these works do not show how one can obtain a converse result, i.e., how to find all extensions, that fulfill some conditions on the Galois structure of the ring of integers. To the best of author’s knowledge the only result obtained in this direction was proved in [By1] and refers only to cyclotomic Lubin-Tate extensions.
0.2. In the examples mentioned above the associated order is also a Hopf order in the group algebra (i.e., it is an order stable under comultiplication). Sev- eral authors are interested in this situation. The extra structure allows to deal with wild extensions as if they were tame (in some sense). This is why in this situation, following Childs, one speaks of “taming wild extensions by Hopf or- ders”. In the paper [By1], Byott proves that the associated order can be a Hopf order only in the case when the different of the extension is generated by an element of the smaller field. The present paper is also dedicated to extensions of this sort. More precisely, Theorem 4.4 of the paper [By1] implies that the order AK/k(OK) can be a Hopf order (in the case when the ringOK iso[G]- indecomposable) only if K/k fulfills the stated condition on the different and OKis free overAK/k(OK). Our main Theorem settles completely the question to determine when the orderAK/k(OK) is a Hopf order. It also describes com- pletely Hopf orders that can be obtained as associated Galois orders. We shall also prove in subsection 3.4 that under the present assumptions ifOK is free over AK/k(OK), then AK/k(OK) is a Hopf order and determine when the in- verse different of an Abelian totally ramifiedp-extension of a complete discrete valuation field is free over its associated order (cf. [By1] Theorem 3.10).
In the first section we study a more general situation. We consider a Galois extension K/k of fraction fields of Dedekind rings, with Galois group G. We prove a formula for the module
CK/k(I1, I2) = Homo(I1, I2),
where I1, I2 are fractional ideals of K (this set also can be defined for ideals that are not G-stable) in Theorem 1.3.1. We introduce two submodules of CK/k(I1, I2):
AK/k(I1, I2) ={f ∈k[G]|f(I1)⊂I2} BK/k(I1, I2) = Homo[G](I1, I2), .
These modules coincide in the case whenK/kis Abelian (cf. Proposition 1.4.2).
We call these modules theassociated modulesfor the pairI1, I2. In caseI1=I2
we call the associated moduleassociated order.
In subsection 1.5 we define a multiplication on the modules of the type CK/k(I1, I2) and show the product of two such modules lies in some third one.
We have also used this multiplication in the study of the decomposability of ideals in extensions of complete discrete valuation fields with inseparable residue field extension (cf. [BV]).
Starting from the second section we consider totally wildly ramified extensions of complete discrete valuation fields with residue field of characteristicpwith the restriction on the different:
DK/k = (δ), δ∈k (∗).
This second section is dedicated to the study of conditions ensuring that the ring of integersOK is free over its associated orderAK/k(OK) orBK/k(OK).
Letn= [K:k].
We prove the following statement.
Proposition. If in the associated order AK/k(OK)(BK/k(OK) resp.) there is an elementξ which maps some (and so, any) elementa∈OK with valuation equal to n−1onto a prime element of the ring OK, then OK ≈AK/k (resp.
BK/k) and besides thatAK/k (resp. BK/k) has a “power” base (in the sense of multiplication ∆∗, cf. the second section), which is constructed explicitly using the elementξ.
The converse to this statement is also proved in the case whenAK/k(OK) (resp.
BK/k(OK)) is indecomposable (cf. the Theorems 2.4.1, 2.4.2). An important part of our reasoning is due to Byott (cf. [By1]).
0.3. The third, fourth and fifth sections are dedicated to proving a more ex- plicit form of the condition of the second section for the Abelian case. The main result of the paper is the following one. We will call an Abelianp-extension of complete discrete valuation fields of characteristic 0almost maximally ramified if its degree divides the different.
Theorem A. Let K/k be an Abelian totally ramified p-extension of p-adic fields, which in case chark= 0 is not almost maximally ramified, and suppose that the different of the extension is generated by an element in the base field (see (*)). Then the following conditions are equivalent:
1. The extension K/k is Kummer for a formal group F, that there exists a formal group F over the ring of integers o of the field k, a finite torsion subgroupT of the formal moduleF(Mo)and a prime elementπ0ofksuch that K=k(x), wherex is a root of the equationP(X) =π0, where
P(X) =Y
t∈T
(X−
F t).
2. The ring OK is isomorphic to the associated order AK/k(OK)as an o[G]- module.
Remark 0.3.1. Besides proving Theorem A we will also construct the element ξ explicitly and so describeAK/k(OK) (cf. the theorems of §2).
Remark 0.3.2. IFkis of characteristic 0 the fact thatOKis indecomposable as ano[G]-module if and only ifK/kis not almost maximally ramified, was proved in [BVZ]. The case of almost maximally ramified extensions is well understood (cf., for example, [Be]). It is obvious, that in the case chark=pthe ringOK is indecomposable as ano[G]-module, because in this case the algebrak[G] is indecomposable.
Remark 0.3.3. In the paper [CM] rings of integers in Kummer extensions for formal groups are also studied as modules over their associated orders. In that paper Kummer extensions are defined with the use of homomorphisms of formal groups. For extensions obtained in this way freeness of the ring of integers over its associated orders is proved. Childs and Moss also use some tensor product to prove their results. Yet their methods seem to be inapplicable for proving inverse results.
Our notion of a Kummer extension for a formal group is essentially equivalent to the one in [CM]. We however only use one formal group and do not impose finiteness restriction on its height.
Besides we consider also the equal characteristic case i.e., chark= chark=p.
Using the methods presented here one can prove that we can actually take the formal groupF in Theorem A of finite height. Yet such a restriction does not seem to be natural.
Remark 0.3.4. Theorem A shows which Hopf orders can be associated to Galois orders for some Abelian extensions. The papers of mathematicians that “tame wild extensions by Hopf orders” do not show that their authors know or guess that such an assertion is valid.
In the third section we study the fields that are Kummer in the sense of part 1 of Theorem A and deduce 2 from 1. In the fourth section we prove that we can suppose the coefficient b1 in ξ =δ−1P
bσσ to be equal to 0. Further, if
b1is equal to 0, then we show that there exists a formal groupF overo, such that forσ ∈Gthebσ, form a torsion subgroup in the formal module F(Mo), i.e.,
bσ+
F bτ =bστ . In the fifth section we prove that ifbσ+
Fbτ is indeed equal tobστ, thenK/kis Kummer for the groupF.
0.4. This paper is the first in a series of papers devoted to associated orders and associated modules. The technique introduced in this paper (especially the mapφand the multiplication∗defined below) turns out to be very useful in studying The Galois structure of ideals. It allows the author to prove in another paper some results about freeness of ideals over their associated orders in extensions that do not fulfill the condition on the different (*). In particular, using Kummer extensions for formal groups we construct a wide variety of extensions in which some ideals are free over their associated orders. These examples are completely new. We also calculate explicitly the Galois structure of all ideals in such extensions. Such a result is very rare. In a large number of cases the necessary and sufficient condition for an ideal to be free over its associated order is found.
The author is deeply grateful to professor S. V. Vostokov for his help and advice.
The work paper is supported by the Russian Fundamental Research Foundation N 01-00-000140.
§1 General results Let
obe a Dedekind ring,
kbe the fraction field of the ring o,
K/k be a Galois extension with Galois group equal toG, D=DK/k be the different of the extensionK/k,
n= [K:k],
tr be the trace operator inK/k.
We also define some associated modules.
ForI1 andI2 G-stable ideals in Kput
BK/k(I1, I2) = Homo[G](I1, I2). ForI1, I2not beingG-stable one can define BK/k(I1, I2) ={f ∈Homo[G](K, K), f(I1)⊂I2}.
For arbitraryI1, I2⊂K we define AK/k(I1, I2) ={f ∈k[G]|f(I1)⊂I2}, CK/k(I1, I2) = Homo(I1, I2).
Obviously, any o-linear map from I1 into I2 can be extended to a k-linear homomorphism fromKintoK. The dimension of Homk(K, K) overkis equal to n2. Now we consider the group algebraK[G]. This algebra acts onK and
the statement in (Bourbaki, algebra,§7, no. 5) implies that a non-zero element ofK[G] corresponds to a non-zero map fromKintoK. Besides the dimension ofK[G] overkis also equal ton2. It follows that any element of Homk(K, K) can be expressed uniquely as an element of K[G]. So we reckon CK/k(I1, I2) being embedded inK[G].
1.1. We consider theG-Galois algebraK⊗kK. It is easily seen that the tensor productK⊗kK is isomorphic to a direct sum ofncopies ofKas ak-algebra.
Being more precise, letKσ, σ∈Gdenote a field, isomorphic toK.
Lemma 1.1.1. There is an isomorphism of K-algebras ψ: K⊗kK→ X
σ∈G
Kσ,
whereψ=P
σψσ andψσ is the projection on the coordinate σ, defined by the equality
ψσ(x⊗y) =xσ(y)∈Kσ.
The proof is quite easy, you can find it, for example, in ”Algebra” of Bourbaki.
Also see 1.2 below.
Now we construct a mapφfrom theG-Galois algebraK⊗kK into the group algebraK[G]:
(1)
φ:K⊗kK→K[G]
α=X
i
xi⊗yi→φ(α) =X
i
xi
X
σ∈G
σ(yi)σ
! .
It is clear thatφ(α) as a function acts onK as follows:
(2) φ(α)(z) =X
i
xitr(yiz), z∈K.
Besides that, the mapφmay be expressed throughψσ in the form
(3) φ(α) =X
σ
ψσ(α)σ, α∈K⊗kK.
Proposition 1.1.2. The mapφis an isomorphism ofk-vector spaces between K⊗kK andK[G].
Cf. the sketch of the proof in Remark 1.2 below.
1.2 Pairings on K⊗kK and K[G].
There is a natural isomorphism of the k-space K and its dual space of k- functionals:
K→Kb a→fa(b) = tr(ab).
We define a pairingh,i⊗ onK⊗kK, that takes its values ink:
(K⊗kK)×(K⊗kK)→k
ha⊗b, c⊗di⊗ →(fa⊗fc)(b⊗d) = (trac)(trbd).
This pairing is correctly defined and is non-degenerate. The second fact is easily proved with the use of dual bases.
We also define a pairing onK[G] that takes its values ink:
K[G]×K[G]→k α=X
σ
aσσ, β=X
σ
bσσ→X
σ
traσbσ=hα, βiK[G].
The pairing h,iK[G] is also non-degenerate because it is a direct sum of n= [K:k] non-degenerate pairings
K×K→k (a, b)→trab.
We check that for any twox, y the following equality is fulfilled:
(4) hx, yi⊗=hφ(x), φ(y)iK[G].
Indeed, it follows from linearity that it is sufficient to prove that for x = a⊗b, y=c⊗d. In that case we have
ha⊗b, c⊗di⊗= tractrbd, and besides that
hφ(a⊗b), φ(c⊗d)iK[G]
=haX
σ
σ(b)σ, cX
σ
σ(d)σiK[G]
=X
σ
tr(acσ(bd)) =X
σ
X
τ
τ(acσ(bd))
=X
τ
X
σ
τ(ac)τ σ(bd) = tractrbd.
and the equality (4) is proved.
Remark 1.2. It follows from (4) that the mapφfrom (1) is an injection. Indeed, let φ(α) be equal to 0 for α ∈ K⊗kK, then hφ(α), φ(β)iK[G] = 0 for any b ∈ K⊗kK. So, according to (4), hα, βi⊗ = 0 for any β ∈ K⊗kK. From the non-degeneracy of the pairing h,i⊗ it follows that α = 0. That implies Ker(φ) = 0.
Using the equality of dimensions we can deduce that φis an isomorphism.
1.3 Modules of homomorphisms for a pair of ideals.
LetI1, I2 be fractional ideals of the fieldK.
Theorem 1.3.1. Letφbe the bijection from (1) and letI1∗=D−1I1−1 (this is dual of I1 for the bilinear trace form on K). For the associated modules the following equality holds:
CK/k(I1, I2) =φ(I2⊗oI1∗).
Proof. First we show thatφ(I2⊗oI1∗)⊂CK/k(I1, I2). If x∈I2, y∈I1∗, then for anyz∈I1we have, according to the definition (2):
φ(x⊗y)(z) =xtr(yz)∈I2,
sincex∈I2 and tr(yz)∈o, which follows from the definition ofI1∗ and of the differentD.
Conversely, let f ∈CK/k(I1, I2). We define a mapθf and show that it is sent ontof byφ. Let:
(5) θf : I2∗⊗oI1→o
x⊗y→tr(xf(y)).
This map is correctly defined sincex∈I2∗=D−1I2−1 andf(y)∈I2, thus tr(xf(y))∈o.
For ao-moduleM we denote byMcthe module ofo-linear functions from M into o. It is clear that
(6) θf ∈(I2∗\⊗oI1).
We identify the idealI2 with theo-moduleIb2∗ via:
I2→Ib2∗ a→fa =X
σ∈G
σ(a)σ.
Obviouslyfa(z) = tr(az) for anyz∈Ib2.
In a completely analogous way we identifyI1∗ withIb1∗
Using these identifications and the fact thatI1andI2∗are projectiveo-modules we obtain an isomorphism
(7) I2⊗oI1∗→(I2∗\⊗oI1) a⊗b→ha,b,
whereha,b(x⊗y) =ha⊗b, x⊗yi⊗= (trax)(trby) for all xin I2∗ andy inI1. The mapθf in (5) lies in(I2∗\⊗oI1) (cf. (6)), and so, according to the isomor- phism (7), it corresponds to an element αf inI2⊗oI1∗, i.e.,
αf =X
i
ai⊗bi, ai∈I2, bi∈I1∗.
Then (7) implies that the functionalhαf, that corresponds to the elementαf, is defined in the following way:
hαf(x⊗y) =hαf, x⊗yi⊗=X
i
traixtrbiy.
On the other hand, from the definition ofθf (cf. (5)) we obtain:
hαf(x⊗y) =θf(x⊗y) = tr(xf(y)).
It follows thatf(y) =P
iaitr(biy). So we have f =φ(X
ai⊗bi),
and for anyf in CK/k(I1, I2) we have found its preimage in I2⊗I1∗, i.e., CK/k(I1, I2)⊂φ(I2⊗oI1∗).
The theorem is proved.
Remark 1.3.2. In the same way as above we can prove, that if we replace the idealsI1, I2 by two arbitrary free o-submodules X and Y ofK of dimension n, then we will obtain the following formula:
CK/k(X, Y) =φ(Y ⊗X),b
whereXb is the dual module toX inKwith respect to the pairingK×K→k defined by the trace tr.
Remark 1.3.3. All elements in CK/k(I1, I2), AK/k(I1, I2), BK/k(I1, I2) have unique extensions to k-linear maps from K to K. To be more precise, if f : I1→I2is ano-homomorphism, then for allx∈Kwe can assumef(x) =αf(a) ifx=αa, α∈k, a∈I1.It is easily seen that the map we obtained in this way is a correctly definedk-linear homomorphism from KintoK.
1.4. Now we compare the modulesAandB. Proposition 1.4.
(8) AK/k(I1, I2) =BK/k(I1, I2) if and only if K/k is an Abelian extension.
Proof. 1. LetK/k be an Abelian extension. To verify the equality (8) in this case, let firstf belong to AK/k(I1, I2), then f is ano-homomorphism fromI1
intoI2. Besides that,f commutes with all elements ofGsinceGis an Abelian group, i.e., σf(a) =f(σ(a)) for all σ ∈Ganda∈I1. So we obtain that f is ano[G]-homomorphism fromI1into I2, thusf ∈BK/k(I1, I2).
For the reverse inclusion, let f belong to BK/k(I1, I2). Then f induces an o[G]-homomorphism from K into K. We take an element x that generates a normal base of the fieldK overk. Then there exists an elementg∈k[G] such that f(x) =g(x), to be more precise, if f(x) = P
σaσσ(x), aσ ∈k then we takeg=P
aσσ. We consider theo[G]-homomorphismgfromKintoK. Since Gis an Abelian group,f(σ(x)) =σ(f(x)) =σ(g(x)) =g(σ(x)) for anyσ∈G.
We obtain thatk-homomorphismsf andgcoincide on the basic elements and so f =g∈k[G] and f(I1)⊂I2, i.e.,f ∈AK/k(I1, I2).
2. Now we suppose thatAK/k(I1, I2) =BK/k(I1, I2) and check thatK/kis an Abelian extension. Indeed, since kAK/k(I1, I2) = k[G], AK/k(I1, I2) contains elements of the form aσ, where a ∈ k∗, for any σ ∈ G. It follows from our assumption that aσ∈BK/k(I1, I2), and soaσ is ano[G]-homomorphism. We obtain thatGcommutes with all elementsσ∈G. Proposition is proved.
Proposition 1.5. If we assume the action of Gon K⊗kK to be diagonal, then
BK/k(I1, I2) =φ((I2⊗oI1∗)G).
Proof. Let α belong to (I2⊗oI1∗)G. We have to show that φ(a) is an o[G]- homomorphism fromI1into I2. This means that
σφ(a)(z) =φ(a)(σ(z)) for all z ∈I1. Let αbe equal to P
ai⊗bi, ai ∈I2, bi ∈I1∗. Then from the definition of φ(cf. (2)) it follows that
φ(a)(σ(z)) =X
i
aitr(biσ(z))
=X
i
aitr(σ−1(bi)σ(z)) =φ X
i
ai⊗σ−1(bi)
! (z).
On the other hand,
σφ(a)(z) =σX
aitr(biz)
=X
(σ(ai) tr(biz)) =φX
σ(ai)⊗bi
(z).
From theG-invariance of the elementαit follows that
φX
ai⊗σ−1(bi)
=φX
σ(ai)⊗σ(σ−1(bi))
=φX
σ(ai)⊗bi
.
Thusφ(α)(σ(z)) =σφ(α)(z) for anyσ ∈G, i.e., φ(α)∈BK/k(I1, I2).
Conversely, let f belong to BK/k(I1, I2), then f ∈ CK/k(I1, I2) and so, ac- cording to Theorem 1.3.1, there is an α ∈ I2⊗I1∗, such that f = φ(α). It remains to check that α is G-invariant. We use the fact that f is an G- homomorphism, i.e., σf(z) = f(σz) for all σ ∈ G and z ∈ I1. We obtain an equalityσφ(α)(z) =φ(α)(σz). By writing the left and the right side of the equality as above we obtain forα=P
ai⊗bi: σφ(α)(z) =φX
σai⊗σ(σ−1bi) (z) φ(α)(σz) =φX
ai⊗σ−1bi
(z).
It follows that
φX
σai⊗σ(σ−1bi)
=φX
ai⊗σ−1bi
.
Now using the fact thatφis a bijection we obtain Xσai⊗σ(σ−1bi) =X
ai⊗σ−1bi. We apply to both sides of the equality the map
1⊗σ: K⊗kK→K⊗kK a⊗b→a⊗σ(b) , that obviously is an homomorphism. We have
Xσai⊗σbi=X ai⊗bi, i.e.,σ(α) =α.
1.6 The multiplication ∗onK[G].
On the algebraK⊗kKthere is a natural multiplication: (a⊗b)·(c⊗d) =ac⊗bd.
Using it and the bijection φwe define a multiplication on K[G]. To be more precise, iff, g∈K[G], then we define
f ∗g=φ(φ−1(f)·φ−1(g))∈K[G]. (9)
Proposition 1.6.1. Iff =P
σaσσ and g=P
σbσσ, then f∗g=X
σ
aσbσσ.
Proof. Let f be equal to φ(α), g be equal to φ(β), where α, β ∈K⊗kK. If α=Pxi⊗yi, β=Puj⊗vj, then from the definition ofφ(cf. (1)) we obtain
φ(α) =X xi
X
σ
σyiσ, φ(β) =X uj
X
σ
σvjσ.
It follows that X
σ
aσbσσ =X
i,j
xiujσ(yivj)σ.
On the other hand,
f ∗g=φ(αβ) =φ
X
i,j
(xiui⊗yivj)
=X
i,j
xiujσ(yivj)σ
and we obtain the proof of Proposition.
Remark 1.6.2. The formula from Proposition 1.6.1 will be used further as an another definition of the multiplication∗.
1.7 Multiplication on associated modules.
Now we consider the multiplication (9) on the different associated modules.
Here we will see appearing the different which we will suppose to be induced from the base field in the following sections.
Proposition 1.7.1. Let f belong to CK/k(I1, I2), and let g belong to CK/k(I3, I4). Then
f ∗g∈CK/k(I1I3D, I2I4).
Proof. It is clear that φ−1(f) and φ−1(g) belong to I2⊗oI1∗ and I4⊗I3∗ re- spectively. So we obtain thatφ−1(f)φ−1(g) lies in the product
(I2⊗oI1∗)(I4⊗oI3∗) =I2I4⊗o(I1∗I3∗)
=I2I4⊗oD−2I1−1I3−1=I2I4⊗o(DI1I3)∗.
So from the Theorem 1.3.1 it follows thatf∗gbelongs toCK/k(I1I3D, I2I4).
Now we study the multiplication∗on the modulesBK/k.
Proposition 1.7.2. Let f belong to BK/k(I1, I2) and let g belong to BK/k(I3, I4), then
f∗g∈BK/k(I1I3D, I2I4).
Proof. SinceBK/k(I, J) is a submodule inCK/k(I, J), from Proposition 1.7 it follows thatf ∗g∈CK/k(I1I3D, I2I4). From Proposition 1.5 we deduce that f and gbelong tophi(K⊗kKG), g∈φ(K⊗kKG). Sof∗g∈φ(K⊗kKG), and this implies that
f∗g∈CK/k(I1I3D, I2I4)∩φ(K⊗kKG) =BK/k(I1I3D, I2I4).
Proposition 1.7.3. Let f belong to AK/k(I1, I2) and LET g belong to AK/k(I3, I4), then
f∗g∈AK/k(I1I3D, I2I4).
Proof. From Proposition 1.7 it follows that
f∗g∈CK/k(I1I3D, I2I4) (10) since AK/k(I1, I2) and AK/k(I3, I4) are submodules OF CK/k(I1, I2) and CK/k(I3, I4) respectively.
From the definition of AK/k it follows that f and g belong to k[G]. So the coefficients off andglie in k, and from Proposition 1.6.1 it follows thatf ∗g also belongs tok[G]. Then (10) implies that
f∗g∈CK/k(I1I3D, I2I4)∩k[G] =AK/k(I1I3D, I2I4), and thus the proposition is proved.
§2 Isomorphism of rings of integers of totally wildly ramified extensions of complete discrete
valuation fields with their associated orders.
2.1. LetK/k be a totally wildly ramified Galois extension of a complete dis- crete valuation field with residue field of characteristicp. LetDbe the different of the extension and let OK be the ring of integers of the fieldK. From this moment and up to the end of the paper we will suppose the condition (*) of the introduction to be fulfilled, i.e., that D = (δ), withδ ∈k. We will write AK/k(OK), BK/k(OK) instead ofAK/k(OK,OK), BK/k(OK,OK).
We denote prime elements of the fieldskandKbyπ0andπ respectively, and their maximal ideals byMo andM.
Proposition 2.1. The modules CK/k(OK), AK/k(OK), BK/k(OK) are o- algebras with a unit with respect to the multiplication
f∆∗g=δf∗g=δ−1φ(φ−1(δf)φ−1(δg)) (cf. (9)). The unit is given by δ−1tr.
The motivation for the above definition is given by Theorem 2.4.1 below.
Proof. Let f and g belong to CK/k(OK), then according to Proposition 1.7, the product f∗g maps the different D into the ringOK. It follows that f∆∗g maps D into D, and so it also maps o into itself since D =δOK. We obtain that ∆∗defines a multiplication on the each of the modules associated toOK. Now we consider the elementδ−1tr and prove that it is the unit for the multi- plication∆∗in each of these modules. It is clear thatδ−1tr mapsOKinto itself and that δ−1tr belongs tok[G], soδ−1tr belongs toAK/k(OK). Besides that, δ−1tr commutes with all elements ofGand soδ−1tr lies inBK/k(OK).
Let nowf belong toK[G], then f =X
σ
aσσ, aσ∈K, δ−1tr =X
σ
δ−1σ.
So, according to proposition 1.6.1,
f∗(δ−1tr) =X
σ
δ−1aσσ and we obtain
f∆∗(δ−1tr) =X aσ=f.
Since AK/k(OK),BK/k(OK) are o-submodules in CK/k(OK), δ−1tr ∈ AK/k(OK), BK/k(OK), δ−1tr is also an identity in AK/k(OK), BK/k(OK) with respect to the multiplication∆∗.
2.2. Let as beforenbe equal to [K:k].
Lemma 2.2.1. Let x be an element of the ring OK whose valuation equals n−1, i.e.,vK(x) =n−1. Then
vk(trx) =vk(δ).
In particular, there is an element a in OK with vK(a) =n−1and such that tra=δ.
Proof. LetMandMobe the maximal ideals ofKandkrespectively. From the definition of the different and surjectivity of the trace operator it follows that tr(M−1D−1) =M−1
o . Moreover any element ofM−1D−1, that does not belong to D−1, has a non-integral trace. So the trace of the element z=π−10 δ−1xis equal to
trz=π0−1δ−1trx=π−10 ε, ε∈o∗.
Thus trx=εδ. Further, if we multiply the elementxbyε−1∈o∗, then we get the elementa.
Lemma 2.2.2. In the ringOK we can choose a basisa0, a1, . . . , an−2, a, where ais as in Lemma 2.2.1, and theai for0≤i≤n−2are such thatvK(ai) =i, and satisfy tr ai= 0.
Proof. The kernel Ker tr(OK) haso-rank equal ton−1. Letx0, . . . , xn−2be an o-basis of Ker tr(OK). Along with the elementathey form ao-base of the ring OK. By elementary operations in Ker tr(OK) we can get from x0, . . . , xn−2
a set of elements with pairwise different valuations. Their valuations have to be less than n−1. Indeed, otherwise by subtracting from the element x0 of valuationn−1 an elementaof the same valuation multiplied by a coefficient in o∗ we can obtain an element ofMn, which is impossible.
2.3. Letabe an element ofOKwith valuation equal ton−1, wheren= [K:k], and let
tra=δ, (12)
whereδ is a generator of the differentDK/k (cf. Lemma 2.2.1).
Proposition 2.3.1. 1. The moduleAK/k(OK)(a) modMn is a subring with an identity inOK modMn (with standard multiplication).
2. The multiplication∆∗ inAK/k(OK)(cf. (11)) induces the standard multipli- cation in the ringAK/k(OK)(a) modMn, i.e.,
f∆∗g(a)≡f(a)g(a) modMn.
Proof. Letf andgbelong toAK/k(OK). Then the preimagesφ−1(δf), φ−1(δg) with respect to the bijectionφbelong toOK⊗oOK, since
D⊗oD−1=δOK⊗oδ−1OK=OK⊗oOK. We prove that
φ−1(δa) =x⊗1 +y, (13)
wherex∈OK andy∈OK⊗M.
Indeed, φ−1(δa) = Pai⊗bi. Ifbi ∈M, then ai⊗bi ∈OK⊗M, otherwise bi=ci+di, whereci∈o,di∈Mand soai⊗bi=ai⊗ci+ai⊗di=cixi⊗1+yi, wherecixi ∈OK, sinceci∈oandyi∈OK⊗M. So (13) follows.
Similarly
φ−1(δg) =x0⊗1 +y0, (14) wherex0∈OK,y0 ∈OK⊗M.
Thus
φ−1(δf)φ−1(δg) = (x⊗1 +y)(x0⊗1 +y0)
=xx0⊗1 +y((x0⊗1) +y0) + (x⊗1)y0
=xx0⊗1 +z, wherez∈OK⊗M.
(15)
We consider the action of the element f ∈ AK/k(OK) on the element awith valuation equal to n−1. From (13) we obtain f = δ−1φ(x⊗1 +y), where x∈OK,y∈OK⊗M. Then from the definition of the mapφwe have:
f(a) =δ−1φ(x⊗1 +y)(a)
=δ−1(φ(x⊗1)(a) +φ(y)(a))
=δ−1xtra+δ−1φ(y)(a).
We show thatδ−1φ(y)(a)∈Mn, i.e.,
f(a) =δ−1xtra+z, wherez∈Mn. (16) Indeed, let y be equal to Pai⊗bi, then from the definition of φ we deduce that
φ(y)(a) =X
aitr(bia).
Moreover trbia∈ DMo, thus δ−1aitr(bia)∈Mn, i.e., z =δ−1φ(y)(a)∈ Mn and we obtain (16).
Our assumptions imply that tra=δ, so
f(a)≡x modMn and similarly
g(a)≡x0 modMn, wherex0 is the element from (14). Then
f(a)g(a)≡xx0 modMn.
On the other hand from the definition of the multiplication f∆∗g (cf. (11)) it follows that
f∆∗g(a) =δ−1φ(φ−1(δf)φ−1(δg))(a).
Using this and keeping in mind (15) and (16) we obtain f∆∗g(a) =δ−1φ(xx0⊗1 +z)(a)≡xx0 modMn. So we have the congruence
f∆∗g(a)≡f(a)g(a) modMn.
Also, the elementδ−1tr inAK/k(OK) gives us an identity element in the ring AK/k(OK)(a) modMn, sinceδ−1tra= 1 (cf. (12)).
Remark 2.3.2. For any other element a0 in the ring OK with valuation equal to n−1 we haveAK/k(OK)(a0)≡εAK/k(OK)(a) modMn, ε∈o∗.
Remark 2.3.3. A similar statement also holds for the module BK/k(OK)(a) modMn.
2.4. Now we formulate the statements which we will begin to prove in the next subsection.
We will investigate the following condition:
in the orderAK/k(OK)(resp. BK/k(OK)) there exists an elementξ such that
ξ(a) =π, (17)
whereπis a prime element of the fieldKandais some element with valuation equal to n−1.
Theorem 2.4.1. If in the ring AK/k(OK) (resp. BK/k(OK)) the condition (17) is fulfilled, then the element ξ generates a “power” basis of AK/k(OK) (BK/k(OK)resp.) over owith respect to the multiplication ∆∗ (cf. 11), i.e.,
AK/k(OK) =hξ0, ξ1, . . . , ξn−1i, whereξ0=δ−1tris the unit andξi =ξ∆∗ξi−1.
Theorem 2.4.2. 1. If for the ring AK/k(OK) (resp. BK/k(OK)) the condi- tion (17) is fulfilled, then the ring OK is a freeAK/k(OK)-module (resp. free BK/k(OK)-module).
2. If the ringOK is a free module over the ringAK/k(OK)(resp. BK/k(OK)) and if moreover the orderAK/k(OK)(resp. BK/k(OK)) is indecomposable (i.e does not contain non-trivial idempotents), then for the ring AK/k(OK) (resp.
BK/k(OK)) the condition (17) and so also the assertions of the theorem 2.4.1 are fulfilled.
Remark 2.4.3. If ξ maps some element with valuation equal to n−1 onto an element with valuation equal to 1, then ξ also maps any other element with valuation equal ton−1 onto an element with valuation equal to 1.
Indeed, if a ∈ K, vk(a) =n−1 and vK(ξ(a)) = 1, then any other element a0 ∈OK, for which vK(a0) =n−1, is equal toεa+b, whereε∈o∗, b∈Mn, so b =π0b0, where π0 is a prime element in k. Besides that b0 ∈OK and we obtainξ(b) =π0ξ(b0) and this impliesvK(ξ(b))≥n.
We also have vK(ξ(εa)) = vK(εξ(a)) = vK(ε) + 1 = 1 i.e., ε ∈ o∗, and so vK(ξ(a0)) = 1.
Remark 2.4.4. Any element ξ in AK/k(OK) (resp. in BK/k(OK)) that fulfills the condition (17) generates a power base of the o-moduleAK/k(OK) (resp.
BK/k(OK)) with respect to the multiplication∆∗.
2.5 Proof of Theorem 2.4.1 and of the first part of Theorem 2.4.2..
We take the element ain the ring OK such thatvK(a) =n−1 and tra=δ (cf. (12)). By assumption we haveξ(a) =π, whereπis a prime element of the field K. We check that
ξi(a)≡πi modMn
K. (18)
Indeed, if i = 0, then for the usual product ξ0(a)δ−1tr(a) = 1. Let further ξi−1(a) be equal to πi−1 modMn
K, then according to Proposition 2.3.1 we have
πi≡ξi−1(a)ξ(a)≡(ξi−1∆∗ξ)(a)≡ξi(a) modMn
K
and the congruence (18) is proved. This equality implies that ξi(a), 0≤i ≤ n−1,generateOK, i.e.,OK =oξ0(a)⊕ · · · ⊕oξn−1(a).Now we show that
AK/k(OK) =hξ0, . . . , ξn−1i.
If there exists an element η ∈AK/k(OK) that does not belong to ao-module hξ0, . . . , ξn−1i, thenη(a) =b∈OK. So we obtain
η(a) =
n−1X
i=0
αiξi(a), αi∈o, i.e.,
(η−X
αiξi)(a) = 0 (19)
Now we show that
η=X
αiξi. (20)
Indeed the spaces kAK/k(OK) andkhξ0, . . . , ξn−1ihave equal dimensions and so coincide. It follows from (19) that
η−X
αiξi∈AK/k(OK)⊂kAK/k(OK) =khξ0, . . . , ξn−1i, and so
η−X αiξi=
n−1X
i=0
α0iξi,
whereα0i∈k. Sinceα0i∈kand the valuations of the elementsξi(a), 0≤i≤n−
1 are pairwise non-congruent modn(cf. (16)), the valuations vK((α0iξi)(a)) are also pairwise non-congruent modn, and so Pα0iξi(a)6= 0 if not all the α0i are equal to 0. This reasoning proves (20).
So we have obtained that the ring OK is a freeAK/k(OK)-module:
OK =AK/k(OK)(a)
and we have proved Theorem 2.4.1 and the first part of Theorem 2.4.2.
Lemma 2.6. Let x be an element of the ring OK such that trx= 0, then for any f ∈AK/k(OK)andg∈BK/k(OK)the following equalities hold:
trf(x) = 0 = trg(x) = 0.
Proof. Iff ∈AK/k(OK), thenf =P
σ∈Gaσσ, aσ∈k, and so trf(x) = tr(X
aσσ(x)) =X
aσtrσ(x) = 0.
If g ∈ BK/k(OK), then trg(x) = g(tr(x)) = g(0) = 0, since g is an o[G]- homomorphism.
2.7. Proof of necessity in Theorem 2.4.2.
LetOK be a freeAK/k(OK)-module and assume the orderAK/k(OK) IS inde- composable. We prove that in the orderAK/k(OK) there exists an elementξ that fulfills the condition (17) of 2.4. We take elementsa0, . . . , an−2such that vK(ai) =iand such that trai= 0 (cf. Lemma 2.2.2 and also [By1]). We take further an elementawith valuation equal ton−1. Letχ: OK →AK/k(OK) be an isomorphism of o[G]-modules, then
AK/k(OK) =hχ(a0), . . . , χ(a)io. Thus, in particular, there existαi, α∈o such that
1 =α0χ(a0) +· · ·+αn−2+αχ(a).
The ringAK/k(OK) is indecomposable by assumption, and so, according to the Krull-Schmidt Theorem, it is a local ring. We obtain that one of theχ(ai) OR χ(a) has to be invertible in the ringAK/k(OK). The elementsχ(ai) cannot be invertible since, according to Lemma 2.6, AK/k(OK)(ai) ∈ Ker trOK. Thus χ(a) is invertible and it follows that AK/k(OK)(a) = OK. We obtain that there exists aξ in AK/k(OK) such thatξ(a) =π. Theorem 2.4.2 is proved.
Remark 2.7. The corresponding reasoning for the ringBK/k(OK) almost lit- erally repeats the one we used above.
§3 Kummer extensions for formal groups.
The proof of sufficiency in Theorem A.
Starting from this section we assume that the extensionK/k is Abelian.
3.1. We denote the valuation onkbyv0. We also denote byv0the valuation onK that coincides withv0onk.
We suppose that the fieldkfulfills the conditions of§2 and thatFis some formal group over the ring o (the coefficients of the series F(X, Y) may, generally speaking, lie, for example, in the ring of integers of some smaller field). On the maximal idealMoof the ringowe introduce a structure of a formalZp-module using the formal groupF by letting forx, y∈Mo andα∈Zp
x+
F y=F(x, y), αx= [α](x).
We denote the Zp-module obtained in this way by F(Mo). Let T be a finite torsion subgroup inF(Mo) and letn= cardT be the cardinality of the group T. Obviously,nis a power ofp.
We construct the following series:
P(X) = Y
t∈T
(X−
Ft). (21)
Remark 3.1.1. The constant term of the series P(X) is equal to zero, the coefficient atXnis invertible ino, and the coefficients at the powers, not equal to n, belong to the idealMo.
Lemma 3.1.2. Let a be a prime element of the field k and K = k(x) be the extension obtained from k by adjoining the roots of the equation P(X) = a, where the seriesP is as in (21). Then the extension K/k is a totally ramified Abelian extension of degree nand the differentD of the extensionK/k is gen- erated by an element of the base field, i.e., D = (δ), δ ∈k. The ramification jumps of the extension K/kare equal to hl=nv0(tl)−1, tl∈T.
Proof. Using the Weierstrass Preparation Lemma we decompose the series P(X)−ainto a product
P(X)−a=cf(X)ε(X),
where ε(X) ∈ o[[X]]∗ is an invertible series (with respect to multiplication), f(X) is A unitary polynomial,c∈o. Then, according to Remark 3.1.1, f(X) is an Eisenstein polynomial of degree n. The series P(X)−a has the same roots (we consider only the roots ofP(X)−awith positive valuation) as the polynomialf(X), and so there are exactlynroots. It is obvious that ifP(x) = a, then
P(x+
F τ) =Y
t∈T
(x−
F t+
Fτ) =Y
t∈T
(x−
Ft) =P(x) =a, τ ∈T and so the roots ofP(X)−aandf(X) are exactly the elementsx+
Fτ, τ ∈T.
Thus we proved that all roots off(X) lie in K and are all distinct. It follows that K/k is a Galois extension. We denote the Galois group of the extension K/k by G. Obviously if σ1, σ2 ∈ G, σ1(x) = x+
F t1, σ2(x) = x+
F t2, then σ2(σ1(x)) = σ2(x+
F t1) = σ2(x) +
F t1 = x+
F t2+
F t1 (as F(X, Y) is defined over o, t1 ∈ T). Since the addition +
F is commutative, the extension K/k is Abelian. We also have that Q
σ(x) =a, vk(a) = 1, and so v0(x) = e(K/k)n . This implies that the extension K/k is totally ramified and thatx is a prime element inK. Now we compute the ramification jumps of the extensionK/k.
LetF(X, Y) =X+Y+P
i,j>0aijXiYj be the formal group law, then we have x−σl(x) =x−(x+
F tl) =tl+ X
i,j>0
aijxitj=tlεt,
where εt is a unit of the ring OK. It follows that the ramification jumps of the extension K/k are equal to nvk(tl)−1. Hence the exponent of the different is equal to P
tl∈T(hl+ 1) = nP
tl∈Tvk(tl) (cf., for example, [Se], Ch. 4, Proposition 4) and sovk(D)≡0 modn.
3.2. Before beginning the proof of Theorem A we prove that the first condition of Theorem A is equivalent to a weaker one.
Proposition 3.2. Let a belong to o, vk(a) = ns + 1, where 0 ≤ s <
mint∈Tvk(t). Then the extension k(x)/k, where x is A root of the equation P(X) =a, has the same properties as the extensions from the first condition of Theorem A.
Proof. We consider the series
Fs(X, Y) =π0−sF(π0sX, π0sY).
It is easily seen that Fs also defines a formal group law and the elements of Ts={π0−st, t∈T}form some torsion subgroup in the formal moduleFs(Mo).
Indeed, if F(X, Y) =PaijXiYj, then Fs(X, Y) =Pπs(i+j−1)0 aijXiYj, and so the coefficients ofFs(X, Y) are integral. Sinceπ−s0 X+
Fs
π0−sY =π0−s(X+
FY), Fs indeed defines an associative and commutative addition. Besides that if u1, u2 ∈ Ts, then u1 +
Fs
u2 = π−s0 (π0su1+
F π0su2) ∈ Ts and so Ts is indeed a subgroup inF(Mo).
Now we compute the seriesPFs(X) for the formal groupFs. We obtain:
PFs(X) =Y
t∈T
(X −
Fs
π0−st) =Y
t
(π0−s(πs0X)−
Fs
π0−st)
=Y
t
π−s0 (π0sX−
F t) =π−sn0 PF(π0sX).
Thus the equation Pf(X) =a is equivalent to PFs(π−s0 X) =aπ0−sn. Besides that v(π−sn0 a) =v(a)−sn, i.e., π0−snais a prime element ink.
Now it remains to note that a root of the equationPf(X) =acan be obtained by multiplication of a root of the equationPFs(Y) =aπ−sn0 byπs0, and so the extensions obtained by adjoining the roots of these equations coincide.
3.3 The proof of Theorem A:1 =⇒ 2.
Let K/k be an extension obtained by adjoining the roots of the equation P(X) = π0. So K =k(x) for some root x. We consider the maximal ideal M⊗ of the tensor product OK⊗OK. Obviously M⊗ =OK⊗M+M⊗OK. We also have Mi
⊗ = Pi
j=0Mj ⊗Mi−j, i > 0, and so it is easily seen that
∩i>0Mi⊗= 0. It follows that we can introduceZp-module structureF(M⊗) on M⊗. We consider the element
α=x⊗1−
F 1⊗x∈M⊗ . We can defineξ in the following way:
ξ=δ−1φ(α). (22)
We check the following properties of the elementξ:
1. ξ∈AK/k(OK)
2. Ify belongs toOK andv(y) =n−1, thenξ(y) is a prime element inK.
For (1):
LetX−
F Y be equal toP
bijXiYj, then the equalities
φ(α) =X
bijφ(xi⊗1i)∗φ(1j⊗xj)
= X
σ∈G, i,j
bijxiσ(xj)σ =X (x−
Fσx)σ=X
tσσ ∈k[G]
follow from the definitions ofφand∗. It also follows from Theorem 1.3.1 that ξ belongs toCK/k(OK). Thusξ ∈k[G]∩CK/k(OK) =AK/k(OK).
For (2):
It easily seen that
(x⊗1)−
F (1⊗x) =x⊗1 +y, (23)
wherey∈M⊗M. Indeed, x⊗1−
F1⊗x=x⊗1 + X
i≥1,j≥0
bijxi⊗xj.
Sincebij∈o,bijxi⊗xj ∈OK⊗Mand we obtain (23).
We can assume that tr(xa) =δ(cf. Remark 2.4.3).
Then, as in Proposition 2.3.1, we have
ξ(a) =δ−1φ(x⊗1 +y)(a)≡x modMn. Sincev(x) = 1, we have proved the property 2.
3.4. Now we construct explicitly a basis of an associated order for extensions that fulfill the condition 1 of Theorem A. We have proved above that we can take the element ξ to be equal to δ−1Ptσσ. Then it follows from Theorem 2.4.1 that
AK/k(OK) =hδ−1X
σ
tiσσ, i= 0, . . . , n−1i.
Now suppose that K is generated by a root of the equationP(X) =b, where vk(b) =sn+ 1, s <minv0(tl). In 3.2 we proved thatK may be generated by a root of the equationPs(X) =bπ0−sn, and so it follows in this case that
AK/k(OK) =hδ−1π−si0 tiσσ, i= 0, . . . , n−1i.
